PHYSICS HIGH SCHOOL

A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical components of its velocity after 5.5sI need shown work !

Answers

Answer 1

Answer:

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

Explanation:

Horizontal Launch

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

$v_y=g.t$

The horizontal component of the velocity is always the same:

$v_x=34 m/s$

The vertical component at t=5.5 s is:

$v_y=9.8*5.5=53.9$

$v_y=53.9\ m/s$

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Find the speed of sound in air that is 30.0°C.

350

314

18

Answers

Answer:

v = 350 m/s

Explanation:

The speed of sound depends on the temperature of surrounding. The relation between the speed of sound and the temperature is given by :

$v=331+0.61T$

Where, T is temperature in Celsius

Here, T = 30 degree Celsius

$v=331+0.61* 30$

v = 349.3 m/s

v = 350 m/s

Therefore, the speed of sound in air at 30 degrees Celsius is 350 m/s.

Answer:

the answer is 350 m/s

Explanation:

If the abosolute of a gas is 600 k, the temperature in degrees Celsius is

Answers

Absolute Zero (Zero Kelvin) = -273.15° Celsius.

Zero° Celsius = 273.15 Kelvin .

600 K = (600 - 273.15°) = 326.85°C.

Jose’s lab instructor gives him a solution of sodium phosphate that is buffered to a pH of 4. Because of an error that he made while calibrating the pH meter, Jose’s measurements of the pH of the sodium phosphate solution are 5.4, 5.4, and 5.4. Which statement best describes his results?

Answers

I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.

Answer: The correct answer is B

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Answers

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A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the magnitude of its acceleration in the water?A. 12 meters/second2
B. 1.5 × 102 meters/second2
C. 78.6 meters/second2
D. 6.5 × 10-3 meters/second2

Answers

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

= zero    - (43 m/s)

=    -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

      = (-43 / 0.28) (m/sec) / sec

      =    153.57... m/s²

      = 1.5... x 10² m/s² .

Answer:

1.5*10^2 meters/seconds^2

Explanation:

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Which of the following shows the abbreviation for a unit of density? (a) lb/mL
(b) g/cm
(c) g/cm3
(d) L/kg

Answers

c) gram/centimeters^3

Answer:

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Explanation:

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